In an article by Wie et al. (1989), for the purpose of calculating the gains of a controller, the equations of motion are approximated by the expression for a damped harmonic oscillation:
$$\ddot{\theta} + d \dot{\theta} + \frac{k}{2} \theta = 0$$
where, of course:
$$d = 2 \zeta \omega _n \quad ; \quad \frac{k}{2} = \omega^2 _n$$
In the design example presented in the article, a settling time of $$T_s = 50 \text{ sec}$$
is assumed. Now, the article states that for a critically damped response, $$\omega _n = 0.158 \text{ rad/s.}$$
Indeed, using this natural frequency to calculate the controller gains yields responses that settle after about 50 seconds:
However, when using the relation between the settling time and the natural frequency for a critically damped system as presented in this answer, one obtaines:
$$
\omega_n \approx \frac{5.8335}{T_s} = \frac{5.8335}{50} = 0.117 \text{ rad/s.}
$$
So the question is, how was the natural frequency calculated in the article?
Best Answer
You need to be careful when using settle time approximations that relate \$\omega_n\$ and \$\zeta\$ to a \$T_s\$.
First, \$T_s\$ is defined as the time where the signal remains within \$\pm x\$% of the final value. Typically the settle limits are \$\pm 2\$% or \$\pm 5\$%, and these limits impact all second order system settle time approximations. Yes, every 2nd order settle time equation is an approximation, and they all differ as a function of \$\zeta\$.
Next, $$\omega_n \approx \frac{5.8335}{T_s},$$ is only a numerical approximation to the normalized unit step response for a critically damped system (i.e., \$\zeta=1\$), which is $$\theta(t)= \left(1-(1+\omega_n t)e^{-\omega_n t}\right)u(t),$$ where \$u(t)\$ is a unit step (aka Heaviside step). The settle time approximation for this is eloquently derived in your "this answer" link using a \$\pm 2\$% window to numerically arrive at \$\omega_n \approx \frac{5.8335}{T_s}\$.
When I look at the graphs of the data you shared, it looks like the 2% settle time for the critically damped trace is closer to 35. This results in \$\omega_n\approx\frac{5.8335}{35}= 0.167\$ rad/sec. A result is closer to the actual \$\omega_n\$ in the article. Better graphs, and clarification on which trace is actually the critically damped one, will improve this estimate.
Sorry, but I do not have free access to your article, so cannot answer your question directly, but most likely the authors knew \$\omega_n\$ directly.